Evaluation of given indefinite integral Evaluate the given integral
$$\int e^x \bigg[\frac{2-x^2}{(1-x)\sqrt{1-x^2}} \bigg]dx$$
I was trying to convert it to $\int e^x (f(x)+f'(x))dx=e^x \cdot f(x)+C$ but did not succeed in algebraic manipulations. Could someone hint me to something so that I could proceed?
 A: We have $$\begin{aligned}
\left[\frac{2-x^2}{(1-x)\sqrt {1-x^2}} \right]  
&= \left[\frac {1-x^2}{(1-x)\sqrt {1-x^2}} + \frac {1}{(1-x)\sqrt {1-x^2}}\right] \\
&= \left[\sqrt {\frac {1+x}{1-x}} + \frac{1}{(1-x)\sqrt {1-x^2}}\right] \\
&= \left[f (x)+g (x)\right] 
\end{aligned}$$
Now take the derivative of $f (x)$ and compare it with $g(x)$. Hope it helps.
A: $$\begin{aligned} 
\frac{2-x^2}{(1-x)\sqrt{1-x^2}} 
&=\frac{1}{(1-x)\sqrt{1-x^2}}+\frac{1-x^2}{(1-x)\sqrt{1-x^2}} \\
&=\frac{1}{(1-x)\sqrt{1-x^2}}+\frac{\sqrt{1-x^2}}{(1-x)} \\
&=\frac{1}{(1-x)\sqrt{1-x^2}}+\frac{\sqrt{1+x}}{\sqrt{1-x}} \\
\end{aligned}$$
The only thing left is to notice that,I'll leave that to you.
$$\left(\frac{\sqrt{1+x}}{\sqrt{1-x}}\right)'=\frac{1}{(1-x)\sqrt{1-x^2}}$$
A: HINT:
As $(1-x)\sqrt{1-x^2}=\sqrt{1+x}(1-x)^{3/2},$
$\displaystyle \int\dfrac1{\sqrt{1+x}}\mathrm dx=2\sqrt{1+x}$
$\displaystyle \int\dfrac1{(1-x)^{3/2}}\mathrm dx=-\dfrac2{\sqrt{1-x}}$
$\mathrm d\left(\dfrac{u}v\right) =\dfrac{u'v-uv'}{v^2}$
So start with $\dfrac{\mathrm d\sqrt{\dfrac{1+x}{1-x}}}{\mathrm dx}$.
